Question

A 0.01 kg spring toy is compressed 0.02 m and released vertically. The toy is measured to reach 0.25 m in the air. Determine the spring constant for the toy.

Answer 1

Answer:

122.5 N/m

Explanation:

According to the law of conservation of energy, if there is no air resistance or frictional forces, the initial elastic potential energy of the spring toy is entirely converted into gravitational potential energy when the toy reaches the highest point.

Therefore, we can write:

$\frac{1}{2}kx^2=mgh$

where the term on the left is the initial elastic potential energy while the term on the right is the gravitational potential energy, and where

k is the spring constant

x = 0.02 m is the compression of the spring

m = 0.01 kg is the mass of the toy

h = 0.25 m is the height reached by the toy

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for k,

$k=\frac{2mgh}{x^2}=\frac{2(0.01)(9.8)(0.25)}{(0.02)^2}=122.5 N/m$